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Measure Theoretic Probability - Video 5 - A Brief History of Integration Theory

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A Brief History of Integration Theory

Fourier Series

Recall the formula for the coefficients of a Fourier series.

\[ c_n = \frac 1 P \int_P f(x)e^{-2\pi i nx/P} \ dx \]

It often happens in applications that this integral is hard to compute. But in some cases we can find a Taylor expansion of the integrand.

\[ f(x) e^{-2\pi inx/P} = \sum a_nx^n \]

It would be wonderful if we could then compute the integral term-by-term. However, this requires swapping the integral with the sum:

\[ \int\sum a_nx^n \ dx = \sum a_n\left( \int x^n \ dx\right) \]

If the sum were finite then there would be no problem — the linearity of integration would tell us that this swaperoo is valid. However, in the infinite case we are not guaranteed this. In fact, we can even re-frame what we need not as the ability to swap the integral and sum, so much as the need to swap the integral and limit. After that, linearity would take care of the rest.

\[ \int\lim_{n\to\infty}\sum_{i=1}^n a_nx^n \ dx = \lim_{n\to\infty}\int\sum_{i=1}^n a_nx^n \]


The Cast of Characters

Fourier himself was concerned with the the problem, seeming to feel that the interchange should be valid. However, later mathematicians demonstrated examples for which it is not valid, and thereby inaugurated a hunt. The hunt was a research project pursued by many mathematicians at the time, to find conditions that one could impose on the function f which would ensure that the interchange of limit and integral was valid.

Some progress was made in this research program. However, the progress was hard-fought. The proofs were extremely long and complicated. Even worse was that the conditions that they found to place on f were extremely strong. So strong that many functions in applications did not meet them.

It was in this context that Lebesgue had a different idea about how to resolve the problem. Rather than place conditions on f, he considered the possibility of re-defining the integral in a more convenient way.


Riemann and Lebesgue

Recall how Riemann integration goes. Suppose we want to integrate f on the interval [a,b].

  1. Partition the domain.

  2. Use the image of f on cells of the partition to determine rectangle heights.

  3. Compute sums of rectangle areas to determine a single approximation.

  4. Take the limit over all approximations, as the number of partition points goes to infinity.

Here’s how Lebesgue integration goes:

  1. Partition the range.

  2. Use the preimage of f on the cells of the partition to determine “pseudo-rectangle” widths.

  3. Compute sums of pseudo-rectangle areas to determine a single approximation.

  4. Take the “limit” over all approximations, as the number of partition points goes to infinity.

Of course the description here is rough, so don’t hold me to my precise wording of Lebesgue integration! We will make this more precise and accurate later.

Pseudo-widths

To help see what is meant by pseudo-widths, consider the function \(f(x)=x^2\) on the interval [-2,2]. Consider the partition of the range [0,4] by the points {0,1,2,4}, for example.

Then the first cell of the partition is [0,1) with preimage \(f^{-1}([0,1)) = [-1,1]\). This has width (and pseudo-width) equal to 2.

The second cell of the partition is [1,2) with preimage \(f^{-1}([1,2)) = [-\sqrt 2,-1)\cup (1,\sqrt 2]\). This has pseudo-width equal to \(-1-(-\sqrt 2) + \sqrt 2-1\).

The third cell is [2,4) with preimage \(f^{-1}([2,4)) = [-4,-\sqrt 2)\cup (\sqrt 2,4]\). This has pseudo-width equal to \(-\sqrt 2-(-4) + 4-\sqrt 2\).


Chaotic Rectangles

If all of the examples were this easy, there would be no real difference between Lebesgue and Riemann integration. However, the preimages of even “nice” continuous, bounded functions can be quite chaotic! The might decompose into a union of intervals. They might be even stranger than that, for all we know!

But in principle what we want is to be able to take any subset of \(\Bbb R\), and assign to it a measure in order to determine these pseudo-widths of pseudo-rectangles.

At this point, our antagonist Vitali enters the scene, to demonstrate that this desire can never be met.

To be continued …